Find the exact value of each expression . Do not use a calculator .

Mathematics

1 answer:

Alex73 [517]3 years ago

8 0

The attached table is one that it is convenient for you to memorize as long as you're taking courses involving angles. For the inverse functions, find the column corresponding to the function, then look for the value in that column. The angle at the beginning of that row is the answer to your question.

8. $\cos^{-1}(0)= 90^{\circ}=\dfrac{\pi}{2}$

9. $\sin^{-1}(1)= 90^{\circ}=\dfrac{\pi}{2}$

10. $\tan^{-1}\left( \dfrac{\sqrt{3}}{3} \right)=30^{\circ}=\dfrac{\pi}{6}$

You might be interested in

OMG PLS HELP WITH THIS IM PANICKING OMG I GOT A F IN MATH MY MOM JUST YELLED AT ME IM CRYING PLS HELP WITH THS- PLSS TELL ME WHA

Marina86 [1]

Answer:

umadddfffd cssssssssssddffg

7 0

2 years ago

A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt

Katena32 [7]

Answer:

$2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}$

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be $y$ . The magnitude of $40\text{ m/s}$ will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is $60^{\circ}$ , we have:

$\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}$ (Recall that $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ )

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation $v_f^2=v_i^2+2a\Delta y$ to solve this problem, where $v_f/v_i$ is final and initial velocity, respectively, $a$ is acceleration, and $\Delta y$ is distance travelled. The only acceleration is acceleration due to gravity, which is approximately $9.8\:\mathrm{m/s^2}$ . However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.